Rearrange Array to minimize difference of sum of squares of odd and even index elements
Given an array arr[] of size N (multiple of 8) where the values in the array will be in the range [a, (a+8*N) -1] (a can be any positive integer), the task is to rearrange the array in a way such that the difference between the sum of squares at odd indices and sum of squares of the elements at even indices is the minimum among all possible rearrangements.
Note: If there are multiple rearrangements return any one of those.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 1 2 4 3 7 8 6 5
Explanation: The difference is 0 as 1 + 42 + 72 + 62 = 102 = 22 + 32 + 82 + 52
Input: arr[] = { 9, 11, 12, 15, 16, 13, 10, 14}
Output: 9 10 12 11 15 16 14 13
Explanation: The difference is 0 as 92 + 122 + 152 + 142 = 102 + 112 + 162 + 132 = 646
Approach: This problem can be solved based on the following mathematical observation:
For any positive integer S, ( S )2 + ( S+3 )2 – 4 = ( S+1 )2 + ( S+2 )2. As N is a multiple of 8 so it can be divided into N/8 groups where the difference of sum of squares of elements at odd and even indices for each group is 0.
For the first four elements keep the sum of squares at odd indices greater and four the next four just the opposite to keep the sum of squares of even indices more. So this group of 8 elements will have difference 0. As the similar is done for all N/8 groups the overall difference will be 0.
The sequence of each group can be like: S, S+1, S+3, S+2, S+6, S+7, S+5, S+4
Follow the below steps to solve this problem:
- Divide the array into groups of size 8.
- Arrange elements in each group as derived from the observation.
- Return the rearranged array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maximum( int arr[], int size)
{
int ma = INT_MIN;
for ( int i = 0; i < size; i++) {
ma = max(ma, arr[i]);
}
return ma;
}
int minimum( int arr[], int size)
{
int mi = INT_MAX;
for ( int i = 0; i < size; i++) {
mi = min(mi, arr[i]);
}
return mi;
}
void print_min( int arr[], int size)
{
int low = minimum(arr, size);
int high = maximum(arr, size);
for ( int i = 0; i < size; i += 4) {
if (i % 8 == 0) {
arr[i] = low;
arr[i + 2] = low + 3;
arr[i + 1] = low + 1;
arr[i + 3] = low + 2;
}
else {
arr[i] = low + 2;
arr[i + 2] = low + 1;
arr[i + 1] = low + 3;
arr[i + 3] = low;
}
low += 4;
}
for ( int i = 0; i < size; i++) {
cout << arr[i] << " " ;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int N = sizeof (arr) / ( sizeof ( int ));
print_min(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
public static int maximum( int arr[], int size)
{
int ma = Integer.MIN_VALUE;
for ( int i = 0 ; i < size; i++) {
ma = Math.max(ma, arr[i]);
}
return ma;
}
public static int minimum( int arr[], int size)
{
int mi = Integer.MAX_VALUE;
for ( int i = 0 ; i < size; i++) {
mi = Math.min(mi, arr[i]);
}
return mi;
}
public static void print_min( int arr[], int size)
{
int low = minimum(arr, size);
int high = maximum(arr, size);
for ( int i = 0 ; i < size; i += 4 ) {
if (i % 8 == 0 ) {
arr[i] = low;
arr[i + 2 ] = low + 3 ;
arr[i + 1 ] = low + 1 ;
arr[i + 3 ] = low + 2 ;
}
else {
arr[i] = low + 2 ;
arr[i + 2 ] = low + 1 ;
arr[i + 1 ] = low + 3 ;
arr[i + 3 ] = low;
}
low += 4 ;
}
for ( int i = 0 ; i < size; i++) {
System.out.print(arr[i] + " " );
}
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 };
int N = arr.length;
print_min(arr, N);
}
}
|
Python3
INT_MIN = - 2147483647 - 1
INT_MAX = 2147483647
def maximum(arr, size):
ma = INT_MIN
for i in range (size):
ma = max (ma, arr[i])
return ma
def minimum(arr, size):
mi = INT_MAX
for i in range (size):
mi = min (mi, arr[i])
return mi
def print_min(arr, size):
low = minimum(arr, size)
high = maximum(arr, size)
for i in range ( 0 ,size, 4 ):
if (i % 8 = = 0 ):
arr[i] = low
arr[i + 2 ] = low + 3
arr[i + 1 ] = low + 1
arr[i + 3 ] = low + 2
else :
arr[i] = low + 2
arr[i + 2 ] = low + 1
arr[i + 1 ] = low + 3
arr[i + 3 ] = low
low + = 4
for i in range (size):
print (arr[i],end = " " )
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ]
N = len (arr)
print_min(arr, N)
|
C#
using System;
class GFG {
static int maximum( int [] arr, int size)
{
int ma = Int32.MinValue;
for ( int i = 0; i < size; i++) {
ma = Math.Max(ma, arr[i]);
}
return ma;
}
static int minimum( int [] arr, int size)
{
int mi = Int32.MaxValue;
for ( int i = 0; i < size; i++) {
mi = Math.Min(mi, arr[i]);
}
return mi;
}
static void print_min( int [] arr, int size)
{
int low = minimum(arr, size);
int high = maximum(arr, size);
for ( int i = 0; i < size; i += 4) {
if (i % 8 == 0) {
arr[i] = low;
arr[i + 2] = low + 3;
arr[i + 1] = low + 1;
arr[i + 3] = low + 2;
}
else {
arr[i] = low + 2;
arr[i + 2] = low + 1;
arr[i + 1] = low + 3;
arr[i + 3] = low;
}
low += 4;
}
for ( int i = 0; i < size; i++) {
Console.Write(arr[i] + " " );
}
}
public static void Main()
{
int [] arr = { 1, 2, 3, 4, 5, 6, 7, 8 };
int N = arr.Length;
print_min(arr, N);
}
}
|
Javascript
<script>
const INT_MIN = -2147483647 - 1;
const INT_MAX = 2147483647;
const maximum = (arr, size) => {
let ma = INT_MIN;
for (let i = 0; i < size; i++) {
ma = Math.max(ma, arr[i]);
}
return ma;
}
const minimum = (arr, size) => {
let mi = INT_MAX;
for (let i = 0; i < size; i++) {
mi = Math.min(mi, arr[i]);
}
return mi;
}
const print_min = (arr, size) => {
let low = minimum(arr, size);
let high = maximum(arr, size);
for (let i = 0; i < size; i += 4) {
if (i % 8 == 0) {
arr[i] = low;
arr[i + 2] = low + 3;
arr[i + 1] = low + 1;
arr[i + 3] = low + 2;
}
else {
arr[i] = low + 2;
arr[i + 2] = low + 1;
arr[i + 1] = low + 3;
arr[i + 3] = low;
}
low += 4;
}
for (let i = 0; i < size; i++) {
document.write(`${arr[i]} `);
}
}
let arr = [1, 2, 3, 4, 5, 6, 7, 8];
let N = arr.length;
print_min(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Another Approach:
- Divide the array into groups of size 8. (a)Since the given array has a length that is a multiple of 8, we can divide it into N/8 groups, where N is the length of the array.
- Arrange elements in each group as derived from the observation. (a)We can use the mathematical observation mentioned in the problem statement:
For any positive integer S, ( S )2 + ( S+3 )2 – 4 = ( S+1 )2 + ( S+2 )2
(b)We can use this formula to derive a sequence of numbers for each group of 8 elements that will keep the difference of sum of squares at odd and even indices zero.
(c)The sequence of each group can be like: S, S+1, S+3, S+2, S+6, S+7, S+5, S+4
(d)We can initialize S to be the minimum value in the array and then increment it by 4 for each group to get the above sequence.
- Return the rearranged array. (a)After arranging each group as described in step 2, we can return the rearranged array.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findMin( int arr[], int size) {
int minElement = INT_MAX;
for ( int i = 0; i < size; i++) {
minElement = min(minElement, arr[i]);
}
return minElement;
}
void rearrangeGroup( int arr[], int startValue) {
arr[0] = startValue;
arr[1] = startValue + 1;
arr[2] = startValue + 3;
arr[3] = startValue + 2;
arr[4] = startValue + 6;
arr[5] = startValue + 7;
arr[6] = startValue + 5;
arr[7] = startValue + 4;
}
void rearrangeArray( int arr[], int size) {
int minElement = findMin(arr, size);
for ( int i = 0; i < size; i += 8) {
if (i % 16 == 0) {
rearrangeGroup(&arr[i], minElement);
} else {
rearrangeGroup(&arr[i], minElement + 2);
}
minElement += 8;
}
}
void printArray( int arr[], int size) {
for ( int i = 0; i < size; i++) {
cout << arr[i] << " " ;
}
cout << endl;
}
int main() {
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int size = sizeof (arr) / sizeof (arr[0]);
rearrangeArray(arr, size);
printArray(arr, size);
return 0;
}
|
Java
import java.util.*;
public class GFG {
public static int findMin( int [] arr, int size) {
int minElement = Integer.MAX_VALUE;
for ( int i = 0 ; i < size; i++) {
minElement = Math.min(minElement, arr[i]);
}
return minElement;
}
public static void rearrangeGroup( int [] arr, int startValue) {
arr[ 0 ] = startValue;
arr[ 1 ] = startValue + 1 ;
arr[ 2 ] = startValue + 3 ;
arr[ 3 ] = startValue + 2 ;
arr[ 4 ] = startValue + 6 ;
arr[ 5 ] = startValue + 7 ;
arr[ 6 ] = startValue + 5 ;
arr[ 7 ] = startValue + 4 ;
}
public static void rearrangeArray( int [] arr, int size) {
int minElement = findMin(arr, size);
for ( int i = 0 ; i < size; i += 8 ) {
if (i % 16 == 0 ) {
rearrangeGroup(arr, minElement);
} else {
rearrangeGroup(arr, minElement + 2 );
}
minElement += 8 ;
}
}
public static void printArray( int [] arr, int size) {
for ( int i = 0 ; i < size; i++) {
System.out.print(arr[i] + " " );
}
System.out.println();
}
public static void main(String[] args) {
int [] arr = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 };
int size = arr.length;
rearrangeArray(arr, size);
printArray(arr, size);
}
}
|
Python3
import sys
def findMin(arr, size):
minElement = sys.maxsize
for i in range (size):
minElement = min (minElement, arr[i])
return minElement
def rearrangeGroup(arr, startValue):
new_arr = [startValue, startValue + 1 , startValue + 3 , startValue +
2 , startValue + 6 , startValue + 7 , startValue + 5 , startValue + 4 ]
return new_arr
def rearrangeArray(arr, size):
minElement = findMin(arr, size)
for i in range ( 0 , size, 8 ):
if i % 16 = = 0 :
new_arr = rearrangeGroup(arr[i:i + 8 ], minElement)
else :
new_arr = rearrangeGroup(arr[i:i + 8 ], minElement + 2 )
arr[i:i + 8 ] = new_arr
minElement + = 8
def printArray(arr, size):
for i in range (size):
print (arr[i], end = " " )
print ()
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ]
size = len (arr)
rearrangeArray(arr, size)
printArray(arr, size)
|
C#
using System;
class GFG
{
static int FindMin( int [] arr, int size) {
int minElement = int .MaxValue;
for ( int i = 0; i < size; i++) {
minElement = Math.Min(minElement, arr[i]);
}
return minElement;
}
static void RearrangeGroup( int [] arr, int startValue) {
arr[0] = startValue;
arr[1] = startValue + 1;
arr[2] = startValue + 3;
arr[3] = startValue + 2;
arr[4] = startValue + 6;
arr[5] = startValue + 7;
arr[6] = startValue + 5;
arr[7] = startValue + 4;
}
static void RearrangeArray( int [] arr, int size) {
int minElement = FindMin(arr, size);
for ( int i = 0; i < size; i += 8) {
if (i % 16 == 0) {
RearrangeGroup(arr, minElement);
} else {
RearrangeGroup(arr, minElement + 2);
}
minElement += 8;
}
}
static void PrintArray( int [] arr, int size) {
for ( int i = 0; i < size; i++) {
Console.Write(arr[i] + " " );
}
Console.WriteLine();
}
static void Main( string [] args) {
int [] arr = { 1, 2, 3, 4, 5, 6, 7, 8 };
int size = arr.Length;
RearrangeArray(arr, size);
PrintArray(arr, size);
}
}
|
Javascript
function findMin(arr) {
let minElement = Infinity;
for (let i = 0; i < arr.length; i++) {
minElement = Math.min(minElement, arr[i]);
}
return minElement;
}
function rearrangeGroup(arr, startValue) {
arr[0] = startValue;
arr[1] = startValue + 1;
arr[2] = startValue + 3;
arr[3] = startValue + 2;
arr[4] = startValue + 6;
arr[5] = startValue + 7;
arr[6] = startValue + 5;
arr[7] = startValue + 4;
}
function rearrangeArray(arr) {
let minElement = findMin(arr);
for (let i = 0; i < arr.length; i += 8) {
if (i % 16 === 0) {
rearrangeGroup(arr, minElement);
} else {
rearrangeGroup(arr, minElement + 2);
}
minElement += 8;
}
}
function printArray(arr) {
console.log(arr.join( ' ' ));
}
let arr = [1, 2, 3, 4, 5, 6, 7, 8];
rearrangeArray(arr);
printArray(arr);
|
Time Complexity: O(N), where N is the size of the array
Auxiliary Space: O(1)
Last Updated :
27 Mar, 2023
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